\(A=\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}+\dfrac{y}{y-x}\left(ĐKXĐ:x\ne\pm y\right)\)

\(A=\left(\dfrac{4xy}{2\left(x-y\right)\left(x+y\right)}+\dfrac{\left(x-y\right)^2}{2\left(x+y\right)\left(x-y\right)}\right):\dfrac{x+y}{2x}+\dfrac{y}{y-x}\)

\(=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}+\dfrac{y}{y-x}\)

\(=\dfrac{x^2+2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}+\dfrac{y}{y-x}\)

\(\dfrac{2x\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)^2}+\dfrac{y}{y-x}=\dfrac{x}{x-y}+\dfrac{y}{y-x}=\dfrac{x}{x-y}-\dfrac{y}{x-y}=\dfrac{x-y}{x-y}=1\)

a: =(xy-2x)-(y^2-2y)

=x(y-2)-y(y-2)

=(x-y)(y-2)

b: =(x^2-2xy+y^2)-(x-y)

=(x-y)^2-(x-y)

=(x-y)(x-y-1)

c: =(x^2-1)-(2xy-2y)

=(x-1)(x+1)-2y(x-1)

=(x-1)(x+1-2y)

d: =(x+3)(x+3-2x+5)

=(x+3)(8-x)

\(a,xy-2x-y^2+2y\)

\(=x\left(y-2\right)-y\left(y-2\right)\)

\(=\left(x-y\right)\left(y-2\right)\)

\(b,x^2-2xy+y^2-x+y\)

\(=\left(x-y\right)^2-\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-1\right)\)

\(c,x^2-1-2xy+2y\)

\(=\left(x-1\right)\left(x+1\right)-2y\left(x-1\right)\)

\(=\left(x-1\right)\left(x+1-2y\right)\)

\(d,\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)

\(=\left(x+3\right)\left(x+3-2x+5\right)\)

\(=\left(x+3\right)\left(-x+8\right)\)

#Urushi

viết sai đề hết rồi

Phân tích đa thức sau thành nhân tử

a) \(2x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow x^2+x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+10x+25\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x\\\left(x+5\right)^2\ge0\forall x\end{cases}}\)

\(\Rightarrow\left(x+y\right)^2+\left(x+5\right)^2\ge0\forall x\)

Vậy đẳng thức xảy ra\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=5\end{cases}}\)

b)\(x^2+3y^2+2xy-2y+1=0\)

\(\Leftrightarrow x^2+y^2+2y^2+2xy-2y+\frac{1}{2}+\frac{1}{2}=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(2y^2-2y+\frac{1}{2}\right)+\frac{1}{2}=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)

Vì \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2\ge0\)

nên \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)

Mà\(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)

nên pt vô nghiệm

a) 2x² + y² + 2xy + 10x + 25 = 0

=> (x² + 2xy + y²) + (x² + 10x + 25) = 0

=> (x + y)² + (x + 5)² = 0 

    <=> \(\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\) <=> \(\hept{\begin{cases}y=-x\\x=-5\end{cases}}\) <=> \(\hept{\begin{cases}y=5\\x=-5\end{cases}}\)

b)c) xem lại đề

a: =(x^2y-x^3)-(9y-9x)

=x^2(y-x)-9(y-x)

=(y-x)(x^2-9)

=(y-x)(x-3)(x+3)

b: \(=\left(x^2-2xy+y^2\right)-4\)

=(x-y)^2-4

=(x-y-2)(x-y+2)

c: \(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

=(x+2+y)(x+2-y)

d: =(x^2-y^2)-(2x+2y)

=(x-y)(x+y)-2(x+y)

=(x+y)(x-y-2)

\(a,x^2y-x^3-9y+9x\)

\(=\left(x^2y-x^3\right)-\left(9y-9x\right)\)

\(=x^2\left(y-x\right)-9\left(y-x\right)\)

\(=\left(y-x\right)\left(x^2-9\right)\)

\(=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)

\(b,x^2-2xy+y^2-4\)

\(=\left(x^2-2xy+y^2\right)-4\)

\(=\left(x-y\right)^2-2^2\)

\(=\left(x-y-2\right)\left(x-y+2\right)\)

\(c,x^2+4x-y^2+4\)

\(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2-y\right)\left(x+2+y\right)\)

\(=\left(x-y+2\right)\left(x+y+2\right)\)

\(d,x^2-y^2-2x-2y\)

\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

#Urushi

m: (x-y)(x^2-2xy+y^2)

=(x-y)*(x-y)^2

=(x-y)^3

=x^3-3x^2y+3xy^2-y^3

n: =-(x^3+x^2y-x-x^2y-xy^2+y)

=-x^3+x+xy^2-y

o: =-(x^3+x^2y^2-x^2-2xy-2y^3+2y)

=-x^3-x^2y^2+x^2+2xy+2y^3-2y

p: (1/2x-1)(2x-3)

=1/2x*2x-1/2x*3-2x+3

=x^2-3/2x-2x+3

=x^2-7/2x+3

q: (x-1/2y)(x-1/2y)

=(x-1/2y)^2

=x^2-xy+1/4y^2

r: (x^2-2x+3)(1/2x-5)

=1/2x^3-5x^2-x^2+10x+3/2x-15

=1/2x^3-6x^2+11,5x-15

bai2 :cmr

a, a^3+b^3=(a+b)^3-3ab.(a+b)

VP= \(\left(a+b\right)^3-3ab\left(a+b\right)\)

=\(a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2=a^3+b^3\)

=VT

b.a^3-b^3=(a-b)^3+3ab,(a-b)

\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)

=\(a^3-3a^2b+ab^2.3-b^3+3a^2b-3ab^2=a^3-b^3\)

=VT

=> ĐPCM

bài 1.

a) = 8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3-(8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3)

= 8x³+4x_²y+2xy_²-4x²y-2xy²-y³ - 8x³+4x²y-2xy²-4x²y+2xy²-y³

=-8x²y-6y³

b) = 27x³-18x²y+12xy²+18x²y-12xy²+8y³-27x³

=8y